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i i i âmainâ 2007/2/16 page 274 i 274 CHAPTER 4 Vector Spaces and setting s = 0 and t = 1 yields the linear combination 2v1 + v4 = 0. (4.5.3) We can solve (4.5.2) for v3 in terms of v1 and v2 , and we can solve (4.5.3) for v4 in terms of v1 . Hence, according to Theorem 4.5.2, we have span{v1 , v2 , v3 , v4 } = span{v1 , v2 }. By Proposition 4.5.7, v1 and v2 are linearly independent, so {v1 , v2 } is the linearly independent set we are seeking. Geometrically, the subspace of R3 spanned by v1 and v2 is a plane, and the vectors v3 and v4 lie in this plane. Linear Dependence and Linear Independence in Rn Let {v1 , v2 , . . . , vk } be a set of vectors in Rn , and let A denote the matrix that has v1 , v2 , . . . , vk as column vectors. Thus, A = [v1 , v2 , . . . , vk ]. (4.5.4) Since each of the given vectors is in Rn , it follows that A has n rows and is therefore an n Ã k matrix. The linear combination c1 v1 + c2 v2 + Â· Â· Â· + ck vk = 0 can be written in matrix form as (see Theorem 2.2.9) Ac = 0, (4.5.5) where A is given in Equation (4.5.4) and c = [c1 c2 . . . ck ]T . Consequently, we can state the following theorem and corollary: Theorem 4.5.14 Let v1 , v2 , . . . , vk be vectors in Rn and A = [v1 , v2 , . . . , vk ]. Then {v1 , v2 , . . . , vk } is linearly dependent if and only if the linear system Ac = 0 has a nontrivial solution. Corollary 4.5.15 Let v1 , v2 , . . . , vk be vectors in Rn and A = [v1 , v2 , . . . , vk ]. 1. If k > n, then {v1 , v2 , . . . , vk } is linearly dependent. 2. If k = n, then {v1 , v2 , . . . , vk } is linearly dependent if and only if det(A) = 0. Proof If k > n, the system (4.5.5) has an inï¬nite number of solutions (see Corollary 2.5.11), hence the vectors are linearly dependent by Theorem 4.5.14. On the other hand, if k = n, the system (4.5.5) is n Ã n, and hence, from Corollary 3.2.5, it has an inï¬nite number of solutions if and only if det(A) = 0. Example 4.5.16 Determine whether the given vectors are linearly dependent or linearly independent in R4 . 1. v1 = (1, 3, â1, 0), v2 = (2, 9, â1, 3), v3 = (4, 5, 6, 11), v4 = (1, â1, 2, 5), v5 = (3, â2, 6, 7). 2. v1 = (1, 4, 1, 7), v2 = (3, â5, 2, 3), v3 = (2, â1, 6, 9), v4 = (â2, 3, 1, 6). i i i i